Question

without using truth table prove that
¬(p↔q) and ¬p↔q are logically equivalent.

Answer 1

Answer on Question #50191 – Math – Discrete Mathematics

without using truth table prove that

$\neg (p \leftrightarrow q)$ and $\neg p \leftrightarrow q$ are logically equivalent.

Solution

Suppose that $\neg (p \leftrightarrow q)$ and $\neg p \leftrightarrow q$ are not logically equivalent.

It means that there exists such $p, q$ that $\overline{p \leftrightarrow q} \neq \bar{p} \leftrightarrow q$ .

Consider two cases:

1) $\overline{p \leftrightarrow q} = 0 \Rightarrow p \leftrightarrow q = 1 \Rightarrow p = q \Rightarrow \bar{p} \neq q \Rightarrow \bar{p} \leftrightarrow q = 0$ . We have obtained that $\overline{p \leftrightarrow q} = \bar{p} \leftrightarrow q$ , because $0 = 0$ . It contradicts the assumption $\overline{p \leftrightarrow q} \neq \bar{p} \leftrightarrow q$ .

2) $\overline{p \leftrightarrow q} = 1 \Rightarrow p \leftrightarrow q = 0 \Rightarrow p \neq q \Rightarrow \bar{p} = q \Rightarrow \bar{p} \leftrightarrow q = 1$ . We have obtained that $\overline{p \leftrightarrow q} = \bar{p} \leftrightarrow q$ , because $1 = 1$ . It contradicts the assumption $\overline{p \leftrightarrow q} \neq \bar{p} \leftrightarrow q$ .

Both cases yield a contradiction. Thus, assumption $\overline{p \leftrightarrow q} \neq \bar{p} \leftrightarrow q$ was wrong, hence $\neg (p \leftrightarrow q)$ and $\neg p \leftrightarrow q$ are logically equivalent.

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